Fundamental experimental objects

1 1 2( ... )na b b bΓ → Decay width = 1/lifetime

1 2 1 2( ... )na a b b bσ → Cross section

Cross section = Transition rate x Number of final states

Initial flux

3

31 (2 )Vd pn

i π=Π

a1v 1V V×

1a 1av

# particles passing through unit area in unit time

# target particles per unit volume

2a

(Lab frame)

The transition rate 4 * ( ) ( ) ( ) ...fi f iT d x x V x xφ φ= − +∫

φi, f → f p

± = e(− ,+ )ip.x 1

2 p0V≡ N

Ve(− ,+ )ip.x

A

B

C

D

Transition rate per unit volume 2

fiTfi TVW =

24 4(2 ) ( )A B C DN N N N

fi C D A BVT p p p pπ δ= − + − − fiM

e.g.

φ f ,i = e(− ,+ )ip.x

244

4

( ) 1 1 1 1(2 )2 2 2 2

C D A Bfi

A B C D

p p p pW

V E E E Eδ

π+ − − ⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

M

Cross section = Transition rate x Number of final states

Initial flux

42 4

32 32

64

1 (2 ) ( )2 (22 ) 2 2C D A B

A B C D

C Dd p p p pE E

dE E

pVV

p d Vπ δπ

σ = + − −Av

M

2

d dQF

σ =M

3 34 4

3 3(2 ) ( )(2 ) 2 (2 ) 2

C DC D A B

C D

d p d pdQ p p p pE E

π δπ π

= + − −Lorentz Invariant Phase space

2 2 2 1/ 2

2 2

4(( . ) )A B

A B A B

F E E

p p m m

=

= −Av

The cross section

The decay rate

212 A

d dQE

Γ = M

1

1

1

334 4

3 3(2 ) ( ... ) ...(2 ) 2 (2 ) 2

n

n

n

BBA B B

B B

d pd pdQ p p p

E Eπ δ

π π= − −

Compton scattering of a π meson

,k λ ', 'k λ

p 'pp k+ p 'pp k−

,k λ ', 'k λ

,k λ ', 'k λ

p 'p

Feynman rules 2

2 2( )

( ) V

V ie A A A

m

eµ µµ

µµ

µ

ψ ψ∂ ∂ +

∂

−

= − + −

=

∂

Klein Gordon

( ' )ie p pλ λ− +

p 'p

,k λ

2 2

ip m− External photon λε

p

2ie

,k λ ', 'k λ

p 'p

γπ γπ→

Compton scattering of a π meson

,k λ ', 'k λ

p 'pp k+ p 'p'p k−

,k λ ', 'k λ ,k λ ', 'k λ

p 'p

22 2

2 2

( ) [ .(2 ) '.(2 ' ')( )

.(2 ' ) '.(2 ') 2 . ']( ')

fiiie p k p k

p k mip k p k i

p k m

ε ε

ε ε ε ε

= − + ++ −

+ − − −− −

iM

3 32 42 4 2

4 6

1 (2 ) ( )2 2 (2 ) 2 2

C DC D A B

A B C D

d p d pVd p p p p VE E V E E

πσ δπ

= + − −Av

M

Compton scattering of a π meson

,k λ ', 'k λ

p 'pp k+ p 'pp k−

,k λ ', 'k λ ,k λ ', 'k λ

p 'p

1

e2M fi = ε.(2 p + k)

i( p + k)2 − m2 ε '.(2 p '+ k ')

+ ε.(2 p '− k ')i

( p '− k ')2 − m2 ε '.(2 p '− k ') − 2iε.ε '

[ ]2 2

22

( . ')1 (1 cos )k

lab m

dd mσ α ε ε

θ⎛ ⎞ =⎜ ⎟Ω⎝ ⎠ + −

22 2 2

0 2

8| . 8.10 3.103total k

d d GeV mbd mπ

σ πασ − − −= = Ω = =

Ω∫

2

/ 12|total k m mkπασ >>

ε.p = ε '.p = 0 Transverse polarisation

Causality?

3

3

'

(2 ) 2( ' )

( ) ( )

p

p

i t t id pF x x i e

D x y D y x

ω

π ω

− − −Δ − = −

= − − −∫ p.(x'-x)

2When ( ) 0, we can perform a Lorentz transformation taking ( ) ( )x y x y x y− < − → − −-m|r| | |...causality preserved (e )m re−−

2No (continuous) transformation possible for (x-y) 0 >

•

•

y

y−

x

0x

x

-imt...and amplitude nonvanishing (e )imte−

2 2-m ( ' ) ( ' )QM : ( '- ) e t tU x x − − −∝ x x

Field theory :

×See Peskin & Schroeder “Quantum Field Theory” p28

Construction of a relativistic field theory

Lagrangian L T V= −

Action

2

1

t

t

S L dt= ∫

Classical path … minimises action

Quantum mechanics … sum over all paths with amplitude /iSe∝

•

•

Lagrangian invariant under all the symmetries of nature

(Nonrelativistic mechanics)

-makes it easy to construct viable theories

3 lagrangian densi, tyL d x= ∫L L

Klein Gordon field ( )xφ

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= } }

T V

Manifestly Lorentz invariant

Lagrangian formulation of the Klein Gordon equation

L = L∫ d 3x, L lagrangian density

Klein Gordon field ( )xφ

L = ∂µφ(x)( )†

∂µφ(x) − m2φ(x)†φ(x)

δS = 0 ⇒

∂L∂φ

− ∂µ ∂L∂(∂µφ)

= 0

Manifestly Lorentz invariant

Euler Lagrange equation

Lagrangian formulation of the Klein Gordon equation

Classical path :

(∂µ∂

µ + m2 )φ = 0 Klein Gordon equation

δL =

∂L∂φ

δφ +∂L

∂(∂µφ)δ (δ ∂µφ)

=

∂L∂φ

− ∂µ ∂L∂(∂µφ)

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥δφ + ∂µ ∂L

∂(∂µφ)δφ

⎛⎝⎜

⎞⎠⎟

surface integral in S..vanishes

δSδφ

= 0

New symmetries

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=

Is invariant under

A symmetry implies a conserved current and charge.

Translation Momentum conservation

Rotation Angular momentum conservation

e.g.

What conservation law does the U(1) invariance imply?

…an Abelian (U(1)) gauge symmetry ( ) ( )ix e xαφ φ→

Noether current

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=

Is invariant under ( ) ( )ix e xαφ φ→ …an Abelian (U(1)) gauge symmetry

†0 ( ) ( )( )

µµδ δφ δ φ φ φ

φ φ∂ ∂= + ∂ + ↔∂ ∂ ∂L L

L=

iαφ i µα φ∂

†( )( ) ( )

i iµ µµ µα φ α φ φ φ

φ φ φ⎡ ⎤⎛ ⎞ ⎛ ⎞∂ ∂ ∂= −∂ + ∂ − ↔⎢ ⎥⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎣ ⎦

L L L

0 (Euler lagrange eqs.)

††0,

2 ( ) ( )iej jµ

µ µ µ µφ φφ φ

⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

L L Noether current

The Klein Gordon current

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=

Is invariant under ( ) ( )ix e xαψ ψ→ …an Abelian (U(1)) gauge symmetry

††0,

2 ( ) ( )iej jµ

µ µ µ µφ φφ φ

⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

L L

( )* *KGj ieµ µ µφ φ φ φ= − ∂ − ∂

This is of the form of the electromagnetic current we used for the KG field

The Klein Gordon current

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L=

Is invariant under φ(x) → eiαφ(x) …an Abelian (U(1)) gauge symmetry

††0,

2 ( ) ( )iej jµ

µ µ µ µφ φφ φ

⎛ ⎞∂ ∂∂ = = −⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠

L L

( )* *KGj ieµ µ µφ φ φ φ= − ∂ − ∂

This is of the form of the electromagnetic current we used for the KG field

3 0Q d x j= ∫ is the associated conserved charge

Suppose we have two fields with different U(1) charges :

1,21,2 1,2( ) ( )i Qx e xαφ φ→

( )( )

† 2 †1 1 1 1

† 2 †2 2 2 2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

x x m x x

x x m x x

µµ

µµ

φ φ φ φ

φ φ φ φ

∂ ∂ −

+ ∂ ∂ −

L=

..no cross terms possible (corresponding to charge conservation)

L = ∂µφ(x)( )†

∂µφ(x) − m2φ(x)†φ(x) + λ φ4+

λ 'M 2 φ

6+ ...

Additional terms

}

Renormalisable D ≤ 4

If M 103GeV , "Effective" Field theory approximately renormalisable

Terms allowed by U(1) symmetry

U(1) local gauge invariance and QED

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= not invariant due to derivatives

( )( ) ( )i x Qx e xαφ φ→

( ) ( ) ( ) ( )i x Q i x Q i x Qe e iQe xα α αµ µ µ µφ φ φ φ α∂ → ∂ = ∂ + ∂

To obtain invariant Lagrangian look for a modified derivative transforming covariantly

Dµφ → eiα ( x )Q Dµφ

U(1) local gauge invariance and QED

( )† 2 †( ) ( ) ( ) ( )x x m x xµµφ φ φ φ∂ ∂ −L= not invariant due to derivatives

( )( ) ( )i x Qx e xαφ φ→

( ) ( ) )( ()( ) (( ) )i x Q i i xx QQ i x Qe e iQe xiQA iQA iQe xαµ µ

α α αµ µ µ µµφ φ φ φ φαφ φ α∂ → ∂ = ∂− − − ∂+ ∂

To obtain invariant Lagrangian look for a modified derivative transforming covariantly

Dµφ → eiα ( x )Q Dµφ

D iQAµ µ µ= ∂ −

A Aµ µ µα→ + ∂Need to introduce a new vector field

( )( ) ( )iQ xx e xαφ φ→

Dµφ → eiα ( x )Q Dµφ

A Aµ µ µα→ + ∂

( )† 2 †( ) ( ) ( ) ( )D x D x m x xµµφ φ φ φ−L= is invariant under local U(1)

Note : D iQAµ µ µ µ∂ → = ∂ − is equivalent to p p eAµ µ µ→ +

universal coupling of electromagnetism follows from local gauge invariance

2 2 2( ) ( )m V where V ie A A e Aµ µ µµ µ µψ ψ∂ ∂ + = − = − ∂ + ∂ −

The Euler lagrange equation give the KG equation:

( )( ) ( )iQ xx e xαφ φ→

Dµφ → eiα ( x )Q Dµφ

A Aµ µ µα→ + ∂

( )† 2 †( ) ( ) ( ) ( )D x D x m x xµµφ φ φ φ−L= is invariant under local U(1)

Note : D iQAµ µ µ µ∂ → = ∂ − is equivalent to p p eAµ µ µ→ +

universal coupling of electromagnetism follows from local gauge invariance

( )†KG 2 † 2( ) ( ) ( ) ( ) ( )KGx x m x x j A O eµ µµ µφ φ φ φ∂ ∂ − − +i.e. L = L =

The electromagnetic Lagrangian

F A Aµν µ ν ν µ= ∂ − ∂

, F F A Aµν µν µ µ µα→ → + ∂

14

EM j AF F µνµν

µµ= − −L

The Euler-Lagrange equations give Maxwell equations !

F jµν νµ∂ =

. , 0

. 0,

tEt

ρ ∂∇ = ∇× + =∂∂∇ = ∇× − =∂

BE E

B B j≡

1 2 3

1 3 2

2 3 1

3 2 1

00

00

E E EE B BE B BE B B

− − −⎛ ⎞⎜ ⎟−⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟−⎝ ⎠

0)A A

µν µ ν

∂ ∂− ∂ =∂ ∂ ∂L L

(

2M A Aµ µ Forbidden by gauge invariance

EM dynamics follows from a local gauge symmetry!!

N.B. εµνρσ∂µFρσ = 0( )

The photon propagator

( )F A A jµν µ ν ν µ νµ µ µ∂ = ∂ ∂ − ∂ ∂ =

• The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations.

2 2 2( ) ( )m V where V ie A A e Aµ µ µµ µ µψ ψ∂ ∂ + = − = − ∂ + ∂ −

The Klein Gordon propagator (reminder)

2 4( ) ( ' ) ( ' )Fm x x x xµµ δ∂ ∂ + Δ − = −

In momentum space:

Δ 'F ( p) = i

− p2 +m2 ± iε

With normalisation convention used in Feynman rules = inverse of momentum space operator multiplied by -i

The photon propagator

( )F A A jµν µ ν ν µ νµ µ µ∂ = ∂ ∂ − ∂ ∂ =

Gauge ambiguity

A Aµ µ µα→ + ∂ 2A Aµ µµ µ α∂ → ∂ + ∂

i.e. with suitable “gauge” choice of α (“ξ” gauge) want to solve

21 1(1 ) ( ) ( (1 ) )A A g A jµ ν ν µ νµ ν µ νµ µ µξ ξ

∂ ∂ − − ∂ ∂ ≡ ∂ − − ∂ ∂ =

In momentum space the photon propagator is

12

2 2

1(1 ) (1 )p pii g p p p g

p pµ νµν µ ν

µν ξξ

− ⎛ ⎞⎛ ⎞− − − = − + −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

(‘t Hooft Feynman gauge ξ=1)

Choose as1

(gauge fixing)

Aµ µξ− ∂

• The propagators determined by terms quadratic in the fields, using the Euler Lagrange equations.